Answer :
To solve the quadratic equation \(x^2 + 6x - 5 = 0\) and express the solutions in the form \(x = m \pm n\), follow these steps:
1. Identify the coefficients:
[tex]\[ a = 1, \quad b = 6, \quad c = -5 \][/tex]
2. Calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac = 6^2 - 4 \cdot 1 \cdot (-5) = 36 + 20 = 56 \][/tex]
3. Find the roots using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
[tex]\[ x = \frac{-6 \pm \sqrt{56}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-6 \pm \sqrt{56}}{2} \][/tex]
[tex]\[ x = \frac{-6 \pm \sqrt{56}}{2} \][/tex]
The two roots are:
[tex]\[ \text{Root 1: } x_1 = \frac{-6 + \sqrt{56}}{2} \approx 0.7416573867739413 \][/tex]
[tex]\[ \text{Root 2: } x_2 = \frac{-6 - \sqrt{56}}{2} \approx -6.741657386773941 \][/tex]
4. Determine \(m\) and \(n\) such that \(x = m \pm n\):
- \(m\) is the average of the two roots:
[tex]\[ m = \frac{x_1 + x_2}{2} = \frac{0.7416573867739413 + (-6.741657386773941)}{2} = \frac{-6}{2} = -3.0 \][/tex]
- \(n\) is the absolute value of the difference between one of the roots and \(m\):
[tex]\[ n = |x_1 - m| = |0.7416573867739413 - (-3.0)| = 3.7416573867739413 \][/tex]
So, the solutions of the equation \(x^2 + 6x - 5 = 0\) are given by:
[tex]\[ x = m \pm n \quad \text{where} \quad m = -3.0 \quad \text{and} \quad n = 3.7416573867739413 \][/tex]
[tex]\(\boxed{-3 \pm 3.7416573867739413}\)[/tex]
1. Identify the coefficients:
[tex]\[ a = 1, \quad b = 6, \quad c = -5 \][/tex]
2. Calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac = 6^2 - 4 \cdot 1 \cdot (-5) = 36 + 20 = 56 \][/tex]
3. Find the roots using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
[tex]\[ x = \frac{-6 \pm \sqrt{56}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-6 \pm \sqrt{56}}{2} \][/tex]
[tex]\[ x = \frac{-6 \pm \sqrt{56}}{2} \][/tex]
The two roots are:
[tex]\[ \text{Root 1: } x_1 = \frac{-6 + \sqrt{56}}{2} \approx 0.7416573867739413 \][/tex]
[tex]\[ \text{Root 2: } x_2 = \frac{-6 - \sqrt{56}}{2} \approx -6.741657386773941 \][/tex]
4. Determine \(m\) and \(n\) such that \(x = m \pm n\):
- \(m\) is the average of the two roots:
[tex]\[ m = \frac{x_1 + x_2}{2} = \frac{0.7416573867739413 + (-6.741657386773941)}{2} = \frac{-6}{2} = -3.0 \][/tex]
- \(n\) is the absolute value of the difference between one of the roots and \(m\):
[tex]\[ n = |x_1 - m| = |0.7416573867739413 - (-3.0)| = 3.7416573867739413 \][/tex]
So, the solutions of the equation \(x^2 + 6x - 5 = 0\) are given by:
[tex]\[ x = m \pm n \quad \text{where} \quad m = -3.0 \quad \text{and} \quad n = 3.7416573867739413 \][/tex]
[tex]\(\boxed{-3 \pm 3.7416573867739413}\)[/tex]